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Calculate \(s^{2}\), the pooled estimator of \(\sigma^{2}\), and provide the degrees of freedom for \(s^{2}\). $$ n_{1}=10, \quad n_{2}=4, \quad s_{1}^{2}=3.4, \quad s_{2}^{2}=4.9 $$

Short Answer

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Question: Calculate the pooled variance and its degrees of freedom when given sample sizes of 10 and 4, and variances of 3.4 and 4.9. Answer: The pooled variance, s^2, is 3.775, and its degrees of freedom is 12.

Step by step solution

01

Calculate the Pooled Variance s^2

To calculate the pooled variance \(s^2\), we will use the following formula: $$ s^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} $$ We are given \(n_1=10, n_2=4, s_1^2=3.4, s_2^2=4.9\). Plug the values into the formula: $$ s^2 = \frac{(10 - 1)(3.4) + (4 - 1)(4.9)}{10 + 4 - 2} $$ Simplify and compute the pooled variance: $$ s^2 = \frac{9(3.4) + 3(4.9)}{12} $$ $$ s^2 = \frac{30.6+14.7}{12} $$ $$ s^2 = \frac{45.3}{12} $$ $$ s^2 = 3.775 $$
02

Calculate the Degrees of Freedom

The degrees of freedom for the pooled variance are given by: $$ d.f. = n_1 + n_2 - 2 $$ We are given \(n_1=10, n_2=4\). Plug the values into the formula: $$ d.f. = 10 + 4 - 2 $$ Compute the degrees of freedom: $$ d.f. = 12 $$ So, the pooled estimator of the variance, \(s^2\), is 3.775, and the degrees of freedom for \(s^2\) is 12.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Degrees of Freedom
Degrees of freedom (d.f.) in statistics represent the number of independent values or quantities which can be assigned to a statistical distribution. In the context of variance, they play a crucial role in accurately estimating the population parameter from a sample statistic.

In the exercise, the degrees of freedom for the pooled variance estimator are calculated using the formula: \( d.f. = n_1 + n_2 - 2 \), where \( n_1 \) and \( n_2 \) are the sample sizes of two groups. The subtraction of 2 accounts for the fact that we have estimated two parameters (in this case, the variances of the two samples) from the data. This adjustment is necessary to ensure an unbiased estimate of the population variance.

Degrees of freedom also affect the shape of the t-distribution, which is used when working with small samples. As the degrees of freedom increase, the t-distribution approaches the normal distribution. In the provided exercise, the degrees of freedom were determined to be 12, which has implications for hypothesis testing and confidence intervals involving the pooled variance.
Variance Estimator
In statistics, a variance estimator is a formula used to calculate the measure of dispersion within a data set. The pooled variance estimator combines variances from two independent samples to obtain a single, more accurate estimate of the population variance, especially when the population variances are assumed to be equal.

The formula for the pooled variance \( s^2 \) is: \[ s^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} \]. This formula weighs the sample variances \( s_1^2 \) and \( s_2^2 \) by their respective degrees of freedom, \( n_1-1 \) and \( n_2-1 \). The result is an estimator that takes into account the sample sizes and variability of each sample.

Understanding how to compute the pooled variance is crucial when performing F-tests or ANOVA (Analysis of Variance) tests, where the homogeneity of variances is a key assumption. Our exercise demonstrates the computation step-by-step, yielding a pooled variance of 3.775. This pooled estimate is more robust than individual sample variances for subsequent inferential statistics.
Statistics Formulas
Statistics formulas are the backbone of data analysis, providing the methods to summarize, describe, and make inferences about the collected data. Formulas such as the one for pooled variance allow us to condense complex data into comprehensible figures that can be used for decision making.

The application of the appropriate formula, however, must be done with understanding of the underlying assumptions and context. When using the pooled variance formula, it's important to be aware that we are assuming the populations from which the samples are taken have the same variance. Without this assumption, a different approach would be needed.

Formulas for calculating other important statistics include the mean, standard deviation, and correlation coefficient. Each of these requires knowledge of the correct degrees of freedom to ensure the validity of the results. In essence, statistics formulas turn raw data into actionable insights, and mastering their use is a fundamental skill for any student of statistics.

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Most popular questions from this chapter

Use the information in Exercise 9 to calculate the observed value of the t statistic for testing the difference between the two population means using paired data. Find a rejection region with \(\alpha=.05\) to state your conclusions. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Does this confirm the results of your hypothesis test? $$\begin{array}{llllll}\hline & \multicolumn{5}{c} {\text { Pairs }} \\\\\cline { 2 - 6 } \text { Population } & 1 & 2 & 3 & 4 & 5 \\\\\hline 1 & 1.3 & 1.6 & 1.1 & 1.4 & 1.7 \\\2 & 1.2 & 1.5 & 1.1 & 1.2 & 1.8 \\\\\hline\end{array}$$

Calculate the number of degrees of freedom for a paired-difference test in Exercises \(2-4,\)with \(n_{1}=n_{2}=\) number of observations in each sample and \(n=\) number of pairs. $$n_{1}=n_{2}=15$$

Drug Potency To properly treat patients, drugs prescribed by physicians must have not only a mean potency value as specified on the drug's container, but also small variation in potency values. A drug manufacturer claims that his drug has a potency of \(5 \pm .1\) milligram per cubic centimeter \((\mathrm{mg} / \mathrm{cc})\). A random sample of four containers gave potency readings equal to \(4.94,5.09,5.03,\) and \(4.90 \mathrm{mg} / \mathrm{cc} .\) a. Do the data present sufficient evidence to indicate that the mean potency differs from \(5 \mathrm{mg} / \mathrm{cc} ?\) b. Do the data present sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer? (HINT: It is sometimes difficult to determine exactly what is meant by limits on potency as specified by a manufacturer. Since he implies that the potency values will fall into the interval \(5 \pm .1 \mathrm{mg} / \mathrm{cc}\) with very high probability - the implication is almost always-let us assume that the range .2 or 4.9 to 5.1 represents \(6 \sigma,\) as suggested by the Empirical Rule.)

Insects hovering in flight expend enormous amounts of energy for their size and weight. The data shown here were taken from a much larger body of data collected by T.M. Casey and colleagues. \({ }^{15}\) They show the wing stroke frequencies (in hertz) for two different species of bees. $$\begin{array}{cc} \hline \text { Species } 1 & \text { Species } 2 \\\\\hline 235 & 180 \\\225 & 169 \\ 190 & 180 \\\188 & 185 \\\& 178 \\\& 182 \\\\\hline\end{array}$$ a. Based on the observed ranges, do you think that a difference exists between the two population variances? b. Use an appropriate test to determine whether a difference exists. c. Explain why a Student's \(t\) -test with a pooled estimator \(s^{2}\) is unsuitable for comparing the mean wing stroke frequencies for the two species of bees.

Cerebral blood flow (CBF) in the brains of healthy people is normally distributed with a mean of 74 . A random sample of 25 stroke patients resulted in an average \(\mathrm{CBF}\) of 69.7 with a standard deviation of \(16.0 .\) a. Test the hypothesis that the standard deviation of CBF measurements for stroke patients is greater than \(\sigma=10\) at the \(\alpha=.05\) level of significance. b. Find bounds on the \(p\) -value for the test. Does this support your conclusion in part a? c. Find a \(95 \%\) confidence interval for \(\sigma^{2}\). Does it include the value \(\sigma^{2}=100 ?\) Does this validate your conclusion in part a?

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